已知等差数列\(\{a _{n} \}\)中，\(2a _{1} +a _{3} =12\)，\(a _{1} +2a _{2} =1+a _{4}\)．
\((1)\)求数列\(\{a _{n} \}\)的通项公式；
\((2)\)记数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n}\)，证明：\( \dfrac {1}{S_{1}+1}+ \dfrac {1}{S_{2}+2}+…+ \dfrac {1}{S_{n}+n} < \dfrac {2}{3}\)．